2559. Count Vowel Strings in Ranges

You are given a 0-indexed array of strings words and a 2D array of integers queries.

Each query queries[i] = [li, ri] asks us to find the number of strings present in the range li to ri (both inclusive) of words that start and end with a vowel.

Return an array ans of size queries.length, where ans[i] is the answer to the ith query.

Note that the vowel letters are 'a', 'e', 'i', 'o', and 'u'.

 Example 1:

Input: words = ["aba","bcb","ece","aa","e"], queries = [[0,2],[1,4],[1,1]]
Output: [2,3,0]
Explanation: The strings starting and ending with a vowel are "aba", "ece", "aa" and "e".
The answer to the query [0,2] is 2 (strings "aba" and "ece").
to query [1,4] is 3 (strings "ece", "aa", "e").
to query [1,1] is 0.
We return [2,3,0].

Example 2:

Input: words = ["a","e","i"], queries = [[0,2],[0,1],[2,2]]
Output: [3,2,1]
Explanation: Every string satisfies the conditions, so we return [3,2,1].

 Constraints:

• 1 <= words.length <= 105
• 1 <= words[i].length <= 40
• words[i] consists only of lowercase English letters.
• sum(words[i].length) <= 3 * 105
• 1 <= queries.length <= 105
• 0 <= li <= ri < words.length

Solution 1: Preprocessing + Binary Search

Solution 2: Prefix Sum

C++

• class Solution {
  public:
      vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
          vector<int> t;
          unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u'};
          for (int i = 0; i < words.size(); ++i) {
              if (vowels.count(words[i][0]) && vowels.count(words[i].back())) {
                  t.push_back(i);
              }
          }
          vector<int> ans;
          for (auto& q : queries) {
              int x = lower_bound(t.begin(), t.end(), q[1] + 1) - lower_bound(t.begin(), t.end(), q[0]);
              ans.push_back(x);
          }
          return ans;
      }
  };

JAVA

• class Solution {
      public int[] vowelStrings(String[] words, int[][] queries) {
          List<Integer> t = new ArrayList<>();
          Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u');
          for (int i = 0; i < words.length; ++i) {
              char a = words[i].charAt(0), b = words[i].charAt(words[i].length() - 1);
              if (vowels.contains(a) && vowels.contains(b)) {
                  t.add(i);
              }
          }
          int[] ans = new int[queries.length];
          for (int i = 0; i < ans.length; ++i) {
              ans[i] = search(t, queries[i][1] + 1) - search(t, queries[i][0]);
          }
          return ans;
      }
  
      private int search(List<Integer> nums, int x) {
          int left = 0, right = nums.size();
          while (left < right) {
              int mid = (left + right) >> 1;
              if (nums.get(mid) >= x) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return left;
      }
  }

PYTHON

• class Solution:
      def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
          vowels = {'a', 'e', 'i', 'o', 'u'}
          t = [i for i, word in enumerate(words) if word[0] in vowels and word[-1] in vowels]
          ans = []
          for l, r in queries:
              count = sum(l <= idx <= r for idx in t)
              ans.append(count)
          return ans